Math question image

A radio station is giving away tickets to a play. They plan to give away tickets to seats that cost $10 or $20. They plan to give away at least 20 tickets, and the total cost of all the tickets can be no more than $300. Make a graph showing how many tickets of each kind can be given away.

Answer:
The maximum number of tickets of each kind that can be given away without exceeding $300 is 10 tickets of each kind.

Explanation:
This problem involves setting up and solving a system of inequalities based on the given constraints: the total cost and the minimum number of tickets of each type. The key concepts involved are linear inequalities, systems of inequalities, and graphical solution methods for linear programming problems.

Steps:

  1. Define variables:
  • Let \( x \) = number of $10 tickets
  • Let \( y \) = number of $20 tickets
  1. Write the constraints:
  • Cost constraint:

Total cost cannot exceed $300:

\[ 10x + 20y \leq 300 \]

  • Minimum tickets constraint:

At least 2 tickets of each kind:

\[ x \geq 2 \]

\[ y \geq 2 \]

  1. Simplify the cost inequality:

Divide the entire inequality by 10:

\[ x + 2y \leq 30 \]

  1. Graphical interpretation:
  • Plot the line \( x + 2y = 30 \) on the coordinate plane.
  • The feasible region is the area below or on this line, and to the right of the lines \( x=2 \) and \( y=2 \).
  1. Find the maximum number of tickets:

Since the goal is to find the maximum number of tickets of each kind without exceeding $300, and considering the constraints \( x \geq 2 \), \( y \geq 2 \), the maximum occurs at the corner point of the feasible region.

  1. Determine the corner points:
  • When \( y=2 \):

\[ x + 2(2) = 30 \Rightarrow x + 4 = 30 \Rightarrow x=26 \]

Check if \( x \geq 2 \): yes, so point \((26, 2)\).

  • When \( x=2 \):

\[ 2 + 2y = 30 \Rightarrow 2y=28 \Rightarrow y=14 \]

Check if \( y \geq 2 \): yes, so point \((2, 14)\).

  • Intersection with axes (if applicable):
  • \( x=0 \):

\[ 0 + 2y \leq 30 \Rightarrow y \leq 15 \]

But \( x \geq 2 \), so ignore \( x=0 \).

  • \( y=0 \):

\[ x + 0 \leq 30 \Rightarrow x \leq 30 \]

But \( y \geq 2 \), so ignore \( y=0 \).

  1. Determine the maximum total tickets:
  • At point \((26, 2)\): total tickets = \( 26 + 2 = 28 \).
  • At point \((2, 14)\): total tickets = \( 2 + 14 = 16 \).

The maximum total tickets are at \((26, 2)\), with 28 tickets.

Summary:

  • The maximum number of tickets of each kind that can be given away without exceeding $300 is **26 tickets of $10 and 2 tickets of $20**.
  • The total number of tickets given away in this case is 28.

Note: The question asks for how many of each kind can be given away, so the answer is 26 tickets of $10 and 2 tickets of $20.

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